Count the number of operations required to reduce the given number

0 0 6 minutes read

Given an integer k and an array op[], in a single operation op[0] will be added to k and then in the second operation k = k + op[1] and so on in a circular manner until k > 0. The task is to print the operation number in which k will be reduced to ? 0. If it impossible to reduce k with the given operations then print -1.
Examples:

Input: op[] = {-60, 10, -100}, k = 100
Output: 3
Operation 1: 100 – 60 = 40
Operation 2: 40 + 10 = 50
Operation 3: 50 – 100 = -50
Input: op[] = {1, 1, -1}, k = 10
Output: -1
Input: op[] = {-60, 65, -1, 14, -25}, k = 100000
Output: 71391

Approach: Count the number of times all the operations can be performed on the number k without actually reducing it to get the result. Then update count = times * n where n is the number of operations. Now, for the remaining operations perform each of the operation one by one and increment count. The first operation when k is reduced to ? 0, print the count.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
 
#include <bits/stdc++.h>
using namespace std;
 
 
int operations(int op[], int n, int k) 
    { 
        int i, count = 0; 
 
        // To store the normalized value 
        // of all the operations 
        int nVal = 0; 
 
        // Minimum possible value for 
        // a series of operations 
        int minimum = INT_MAX; 
        for (i = 0; i < n; i++) 
        { 
            nVal += op[i]; 
            minimum  = min(minimum , nVal); 
 
            // If k can be reduced with 
            // first (i + 1) operations 
            if ((k + nVal) <= 0) 
                return (i + 1); 
        } 
 
        // Impossible to reduce k 
        if (nVal >= 0) 
            return -1; 
 
        // Number of times all the operations 
        // can be performed on k without 
        // reducing it to <= 0 
        int times = (k - abs(minimum )) / abs(nVal); 
 
        // Perform operations 
        k = (k - (times * abs(nVal))); 
        count = (times * n); 
 
        // Final check 
        while (k > 0) { 
            for (i = 0; i < n; i++) { 
                k = k + op[i]; 
                count++; 
                if (k <= 0) 
                    break; 
            } 
        } 
 
        return count; 
    } 
 
// Driver code
int main() {
     
        int op[] = { -60, 65, -1, 14, -25 }; 
        int n = sizeof(op)/sizeof(op[0]); 
        int k = 100000; 
 
        cout << operations(op, n, k) << endl; 
}
// This code is contributed by Ryuga

Java

// Java implementation of the approach
class GFG {
 
    static int operations(int op[], int n, int k)
    {
        int i, count = 0;
 
        // To store the normalized value
        // of all the operations
        int nVal = 0;
 
        // Minimum possible value for
        // a series of operations
        int min = Integer.MAX_VALUE;
        for (i = 0; i < n; i++) {
            nVal += op[i];
            min = Math.min(min, nVal);
 
            // If k can be reduced with
            // first (i + 1) operations
            if ((k + nVal) <= 0)
                return (i + 1);
        }
 
        // Impossible to reduce k
        if (nVal >= 0)
            return -1;
 
        // Number of times all the operations
        // can be performed on k without
        // reducing it to <= 0
        int times = (k - Math.abs(min)) / Math.abs(nVal);
 
        // Perform operations
        k = (k - (times * Math.abs(nVal)));
        count = (times * n);
 
        // Final check
        while (k > 0) {
            for (i = 0; i < n; i++) {
                k = k + op[i];
                count++;
                if (k <= 0)
                    break;
            }
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int op[] = { -60, 65, -1, 14, -25 };
        int n = op.length;
        int k = 100000;
 
        System.out.print(operations(op, n, k));
    }
}

Python3

# Python3 implementation of the approach 
def operations(op, n, k):
 
    i, count = 0, 0
 
    # To store the normalized value
    # of all the operations
    nVal = 0
 
    # Minimum possible value for
    # a series of operations
    minimum = 10**9
    for i in range(n):
        nVal += op[i]
        minimum = min(minimum , nVal)
 
        # If k can be reduced with
        # first (i + 1) operations
        if ((k + nVal) <= 0):
            return (i + 1)
 
    # Impossible to reduce k
    if (nVal >= 0):
        return -1
 
    # Number of times all the operations
    # can be performed on k without
    # reducing it to <= 0
    times = (k - abs(minimum )) // abs(nVal)
 
    # Perform operations
    k = (k - (times * abs(nVal)))
    count = (times * n)
 
    # Final check
    while (k > 0):
        for i in range(n):
            k = k + op[i]
            count += 1
            if (k <= 0):
                break
 
    return count
 
# Driver code
op = [-60, 65, -1, 14, -25]
n = len(op)
k = 100000
 
print(operations(op, n, k))
 
# This code is contributed 
# by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG 
{
 
    static int operations(int []op, int n, int k)
    {
        int i, count = 0;
 
        // To store the normalized value
        // of all the operations
        int nVal = 0;
 
        // Minimum possible value for
        // a series of operations
        int min = int.MaxValue;
        for (i = 0; i < n; i++)
        {
            nVal += op[i];
            min = Math.Min(min, nVal);
 
            // If k can be reduced with
            // first (i + 1) operations
            if ((k + nVal) <= 0)
                return (i + 1);
        }
 
        // Impossible to reduce k
        if (nVal >= 0)
            return -1;
 
        // Number of times all the operations
        // can be performed on k without
        // reducing it to <= 0
        int times = (k - Math.Abs(min)) / Math.Abs(nVal);
 
        // Perform operations
        k = (k - (times * Math.Abs(nVal)));
        count = (times * n);
 
        // Final check
        while (k > 0) 
        {
            for (i = 0; i < n; i++) 
            {
                k = k + op[i];
                count++;
                if (k <= 0)
                    break;
            }
        }
 
        return count;
    }
 
    // Driver code
    static void Main()
    {
        int []op = { -60, 65, -1, 14, -25 };
        int n = op.Length;
        int k = 100000;
 
        Console.WriteLine(operations(op, n, k));
    }
}
 
// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach 
function operations($op, $n, $k) 
{ 
    $count = 0; 
 
    // To store the normalized value 
    // of all the operations 
    $nVal = 0; 
 
    // Minimum possible value for 
    // a series of operations 
    $minimum = PHP_INT_MAX; 
    for ($i = 0; $i < $n; $i++) 
    { 
        $nVal += $op[$i]; 
        $minimum = min($minimum , $nVal); 
 
        // If k can be reduced with 
        // first (i + 1) operations 
        if (($k + $nVal) <= 0) 
            return ($i + 1); 
    } 
 
    // Impossible to reduce k 
    if ($nVal >= 0) 
        return -1; 
 
    // Number of times all the operations 
    // can be performed on k without 
    // reducing it to <= 0 
    $times = round(($k - abs($minimum )) / 
                         abs($nVal)); 
 
    // Perform operations 
    $k = ($k - ($times * abs($nVal))); 
    $count = ($times * $n); 
 
    // Final check 
    while ($k > 0) 
    { 
        for ($i = 0; $i < $n; $i++) 
        { 
            $k = $k + $op[$i]; 
            $count++; 
            if ($k <= 0) 
                break; 
        } 
    } 
 
    return $count; 
} 
 
// Driver code
$op = array(-60, 65, -1, 14, -25 ); 
$n = sizeof($op); 
$k = 100000; 
 
echo operations($op, $n, $k); 
 
// This code is contributed by ihritik
?>

Javascript

<script>
// Javascript implementation of the approach
 
function operations(op,n,k)
{
    let i, count = 0;
   
        // To store the normalized value
        // of all the operations
        let nVal = 0;
   
        // Minimum possible value for
        // a series of operations
        let min = Number.MAX_VALUE;
        for (i = 0; i < n; i++) {
            nVal += op[i];
            min = Math.min(min, nVal);
   
            // If k can be reduced with
            // first (i + 1) operations
            if ((k + nVal) <= 0)
                return (i + 1);
        }
   
        // Impossible to reduce k
        if (nVal >= 0)
            return -1;
   
        // Number of times all the operations
        // can be performed on k without
        // reducing it to <= 0
        let times = Math.floor((k - Math.abs(min)) / Math.abs(nVal));
   
        // Perform operations
        k = (k - (times * Math.abs(nVal)));
        count = (times * n);
   
        // Final check
        while (k > 0) {
            for (i = 0; i < n; i++) {
                k = k + op[i];
                count++;
                if (k <= 0)
                    break;
            }
        }
   
        return count;
}
 
    // Driver code
    let op=[-60, 65, -1, 14, -25];
    let n = op.length;
    let k = 100000;
    document.write(operations(op, n, k));
     
// This code is contributed by unknown2108.
</script>

Output:

Time Complexity: O(n*k)

Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!